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Q.An insurance company has insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of meeting with an accident is 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

Tripura TbseHigher Secondary (+2 Stage) Examination 2025Subjective· 4mImportance★★★★★
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This is a Bayes'-theorem problem: weight each group's accident probability by its size, then find what fraction of the total "expected accidents" comes from scooter drivers.

Let SS, CC, TT denote scooter, car, truck drivers. Total insured =2000+4000+6000=12000=2000+4000+6000=12000.

P(S)=200012000=16,P(C)=400012000=13,P(T)=600012000=12.P(S)=\dfrac{2000}{12000}=\dfrac16,\quad P(C)=\dfrac{4000}{12000}=\dfrac13,\quad P(T)=\dfrac{6000}{12000}=\dfrac12.

Accident probabilities given: P(A∣S)=0.01P(A\mid S)=0.01, P(A∣C)=0.03P(A\mid C)=0.03, P(A∣T)=0.15P(A\mid T)=0.15.

By the total probability theorem:

P(A)=P(S)P(A∣S)+P(C)P(A∣C)+P(T)P(A∣T).P(A)=P(S)P(A\mid S)+P(C)P(A\mid C)+P(T)P(A\mid T).

It's cleanest to work directly with the raw counts (equivalent, and avoids fraction clutter): expected number of accidents from each group is …

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